Question

The temperature gradient in the earth’s crust is ${32}^{\circ }C$ per km and the mean conductivity of the rock is 0.008 CGS units. Considering the radius of the earth is 6000km, the loss of heat by earth everyday is about(in calories):
$\text{A.}{10}^{16}$
$\text{B.}{10}^{18}$
$\text{C.}{10}^{27}$
$\text{D.}{10}^{37}$

Answer 1

Hint: Temperature gradient is the rate of change of temperature on moving some distance along the path along which temperature is changing. Since we have a surface whose ends are at different temperatures, hence we have the flow of heat through the surface, continuously. This is a typical case of heat conduction through a spherical surface.

Formula used:
${\displaystyle \frac{dQ}{dt}}=KA{\displaystyle \frac{dT}{dx}}$

Complete answer:
In the formula ${\displaystyle \frac{dQ}{dt}}=KA{\displaystyle \frac{dT}{dx}}$, K is the average conductivity and is given to be 0.008 CGS units for earth. Also the area (A) for the earth is to be taken to the surface area. Thus $A=4\pi R^2$, where R is the radius of earth and is equal to 6400 km and for CGS units, we are supposed to take it in ‘cm’ i.e. CGS units of length. Thus $R=6400\times {10}^{5}cm$ The term ${\displaystyle \frac{dT}{dx}}$ is called the temperature gradient and is equal to ${32}^{\circ }C/km$. Hence in CGS units, ${32}^{\circ }$ $\times$ ${10}^{-5}$ ${}^{\circ }$ $C/cm$.
Now, putting the values in the equation ${\displaystyle \frac{dQ}{dt}}=KA{\displaystyle \frac{dT}{dx}}$, we get
${\displaystyle \frac{dQ}{dt}}=0.008\times 4\pi (6400\times {10}^5{)}^2\times 32\times {10}^{-5}=1.3\times {10}^{13}cal/s$
Now, we are asked the total heat lost in 1 day. Total seconds in 1 day = $24hours=24\times 60minutes=24\times 60\times 60seconds=86400seconds.$
Hence total heat loss in 86400 seconds = $1.3\times {10}^{13}\times 86400\approx 1.1\times {10}^{18}calories$
Hence, option B is correct.

Note: Since we knew the temperature gradient of the earth’s surface, hence we could directly apply the law ${\displaystyle \frac{dQ}{dt}}=KA{\displaystyle \frac{dT}{dx}}$ over the surface as the heat is being loose from the surface. Students should take care of the CGS and S.I. units and their respective conversions. Otherwise even if we are conceptually correct, our answer will be wrong. Students should avoid these types of mistakes. Also the simpler data like the radius of earth is needed to be remembered. This type of data (mass, radius of earth) is usually don’t given in the exams and we are supposed to remember them.